Question

. Water at 100 oC is taken off the stove and allowed to cool for 10 minutes. In this 10
minutes, the temperature decreased to 65oC. Given that the room temperature is 21oC,
find the temperature of the water in degrees Celsius after an additional 5 minutes
waiting time and on condition that the Newton’s Law of Cooling is not violated.

Answer 1

Answer:

Water at 100 oC is taken off the stove and allowed

Answer 2

Answer: 80.32°C

Explanation:

Given:  Water at 100°C is taken off the stove and allowed to cool for 10 minutes

In these 10 minutes, the temperature decreased to 65°C. Given that the room temperature is 21°C.

To Find: The temperature of the water in degrees Celsius after an additional 5 minutes

Solution: From Newton's law of cooling:

-dθ/dt ∝ (θ - θs)

where -dθ/dt is rate of cooling

θ = average temperature of body

θs = average temperature of surrounding

Here, 35/10 = k((65 + 100)/2 - 21)

⇒ 3.5 = k × 123/2

⇒ k = 7/123    -----(i)

and (100 - θ)/5 = k((100 + θ)/2 - 21)   -----(ii)

Putting k = 7/123 from equation i to equation ii we get:-

(100 - θ)/5 = 7/123((100 + θ)/2 - 21)

22570 = 281θ

θ = 22570/281

θ = 80.32°C

Hence, the required temperature of the water after an additional 5 minutes is 80.32°C.

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